Mass Ma lies on top of mass Mb, as shown. Assume Mb > Ma. The two blocks are pulled from rest by a massless rope passing over a pulley. The pulley is accelerated at rate A. Block Mb slides on the table without friction, but there is a constant friction force f between Ma and Mb due to their relative motion. Find the tension in the rope.
The Attempt at a Solution
Newton's second law in the y direction for both masses:
[itex]N_1 - M_a g = 0[/itex]
[itex]N_2 - N_1 - M_b g = 0[/itex]
In the x-direction:
[itex]T_a - F_a = M_a a_a[/itex]
[itex]T_b + F_a = M_b a_b[/itex]
[itex]F_a = \mu M_a g[/itex]
Ok, so I suppose we can assume the tension in the rope is the same at every point since it's massless. So [itex]T_a = T_b = T[/itex]. Now, the thing is, friction can't cause the blocks to move, it can only oppose motion. So maybe the friction here has to be static friction and we assume the blocks are not moving relative to each other? Otherwise, one of the blocks would have to be moving opposite the direction in which we apply the force, which seems wrong. I'm really not sure where to go with this problem.